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  • The radius of the circle passing through the foci of the ellipse <img alt="\mathrm{\frac{x^2}{16}+\frac{y^2}{9}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%

The radius of the circle passing through the foci of the ellipse \mathrm{\frac{x^2}{16}+\frac{y^2}{9}=1} and having its centre at \mathrm{(0,3)} is
 

Option: 1

4


Option: 2

3


Option: 3

\sqrt12


Option: 4

7/12


Answers (1)

best_answer

\mathrm{\frac{x^2}{16}+\frac{y^2}{9}=1}

\mathrm{As \: b^2=a^2\left(1-e^2\right)}

\mathrm{\Rightarrow \quad 9=16\left(1-e^2\right)}

\mathrm{Or \, \, e^2=1-\frac{9}{16}=\frac{7}{16} \Rightarrow e=\frac{\sqrt{7}}{4}}

\mathrm{\therefore \quad S \ is\: (a e, 0)=\left(4 \frac{\sqrt{7}}{4}, 0\right)=(\sqrt{7}, 0)}

The centre of the required circle is at \mathrm{ C(0,3).}

Radius of the circle \mathrm{=C S=\sqrt{7+9}=4.}

Hence option 1 is correct.
 

Posted by

Irshad Anwar

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